There is in fact a perfect family of such sets, given as the branches of a certain perfect binary tree. (By a perfect binary tree I mean a rooted tree in which each element has 1 or 2 successors, such that there is no isolated branch, i.e., above each node of the tree there is at least one "splitting" node, i.e., one with two successors.)

(Via characteristic functions, this family can be viewed as a subset of $\{0,1\}^{\mathbb N}$, the set of all infinite $\{0,1\}$-sequences. Both the space $\{0,1\}^{\mathbb N}$ and the family I constructed are naturally homeomorphic to the Cantor set, and in particular have the same cardinality as the real numbers.)

The root of the tree is the number 1. It has two successors, 2 and 3. Above 2 there is a sequence 4,6,8,...,14 of nonsplitting nodes; note that the sum of their inverses is more than 1. Also above 3 there is a sequence 5,7,9,...,15 of nonsplitting nodes, again with sum of inverses greater than 1.

Above 14 we add a splitting node 16 with successors 18 and 20, and above 15 we add a splitting node 17 with successors 19 and 21. Above each of nodes 18,19,20,21 we add a finite arithmetic sequence of nonsplitting nodes with difference 4 (say: 22,26,... etc above 18; 23,27,... above 19, etc.) in such a way that each respective sum of reciprocals is greater than 1. These arithmetic sequences will be disjoint.

Above each of these 4 arithmetic sequences we add a splitting node; above those we attach 8 disjoint arithmetic sequences with difference 8 which are again so long that their sum of reciprocals is each greater than 1. Etc.

Clearly any two branches of this tree will be almost disjoint sets, and along any branch the sum of reciprocals diverges.

There is in fact a perfect family of such sets, given as the branches of a certain perfect binary tree. (By a perfect binary tree I mean a rooted tree in which each element has 1 or 2 successors, such that there is no isolated branch, i.e., above each node of the tree there is at least one "splitting" node, i.e., one with two successors.)

(Via characteristic functions, this family can be viewed as a subset of $\{0,1\}^{\mathbb N}$, the set of all infinite $\{0,1\}$-sequences. Both the space $\{0,1\}^{\mathbb N}$ and the family I constructed are naturally homeomorphic to the Cantor set, and in particular have the same cardinality as the real numbers.)

The root of the tree is the number 1. It has two successors, 2 and 3. Above 2 there is a sequence 4,6,8,...,14 of nonsplitting nodes; note that the sum of their inverses is more than 1. Also above 3 there is a sequence 5,7,9,...,15 of nonsplitting nodes, again with sum of inverses greater than 1.

Above 14 we add a splitting node 16 with successors 18 and 20, and above 15 we add a splitting node 17 with successors 19 and 21. Above each of nodes 18,19,20,21 we add a finite arithmetic sequence of nonsplitting nodes with difference 4 (say: 22,26,... etc above 18; 23,27,... above 19, etc.) in such a way that each respective sum of reciprocals is greater than 1. These arithmetic sequences will be disjoint.

Above each of these 4 arithmetic sequences we add a splitting node; above those we attach 8 disjoint arithmetic sequences with difference 8 which are again so long that their sum of reciprocals is each greater than 1. Etc.

Clearly any two branches of this tree will be almost disjoint sets, and along any branch the sum of reciprocals diverges.

An alternative (perhaps more transparent) formulation: Let $2^{<\omega}$ be the set of all finite $\{0,1\}$-sequences; this is a countable set. Find a countable family of pairwise disjoint finite sets $(A_s:s\in 2^{<\omega})$ such that $\sum_{n\in A_s}\frac1n \ge 1$ for each $s\in 2^{<\omega}$. Now every infinite $\{0,1\}$-sequence $f$ will define a set $A_f:=\bigcup _{s\vartriangleleft f} A_s$. (Here, $s\vartriangleleft f$ means that the finite sequence $s$ is an initial segment of $f$.)

These sets $A_f$ (for each $f:\mathbb N \to \{0,1\}$) will now be almost disjoint, and the harmonic series diverges along each of them.

A similar construction can be done for any divergent series.

There is in fact a perfect family of such sets, given as the branches of a certain perfect binary tree. (By a perfect binary tree I mean a rooted tree in which each element has 1 or 2 successors, such that there is no isolated branch, i.e., above each node of the tree there is at least one "splitting" node, i.e., one with two successors.)

The root is the number 1. It has two successors, 2 and 3. Above 2 there is a sequence 4,6,8,...,14 of nonsplitting nodes; note that the sum of their inverses is more than 1. Also above 3 there is a sequence 5,7,9,...,15 of nonsplitting nodes, again with sum of inverses greater than 1.

Above 14 we add a splitting node 16 with successors 18 and 20, and above 15 we add a splitting node 17 with successors 19 and 21. Above each of nodes 18,19,20,21 we add a finite arithmetic sequence of nonsplitting nodes with difference 4 (say: 22,26,... etc above 18; 23,27,... above 19, etc.) in such a way that each respective sum of reciprocals is greater than 1. These arithmetic sequences will be disjoint.

Above each of these 4 arithmetic sequences we add a splitting node; above those we attach 8 disjoint arithmetic sequences with difference 8 which are again so long that their sum of reciprocals is each greater than 1. Etc.

There is in fact a perfect family of such sets, given as the branches of a certain perfect binary tree. (By a perfect binary tree I mean a rooted tree in which each element has 1 or 2 successors, such that there is no isolated branch, i.e., above each node of the tree there is at least one "splitting" node, i.e., one with two successors.)

(Via characteristic functions, this family can be viewed as a subset of $\{0,1\}^{\mathbb N}$, the set of all infinite $\{0,1\}$-sequences. Both the space $\{0,1\}^{\mathbb N}$ and the family I constructed are naturally homeomorphic to the Cantor set, and in particular have the same cardinality as the real numbers.)

The root of the tree is the number 1. It has two successors, 2 and 3. Above 2 there is a sequence 4,6,8,...,14 of nonsplitting nodes; note that the sum of their inverses is more than 1. Also above 3 there is a sequence 5,7,9,...,15 of nonsplitting nodes, again with sum of inverses greater than 1.

Above 14 we add a splitting node 16 with successors 18 and 20, and above 15 we add a splitting node 17 with successors 19 and 21. Above each of nodes 18,19,20,21 we add a finite arithmetic sequence of nonsplitting nodes with difference 4 (say: 22,26,... etc above 18; 23,27,... above 19, etc.) in such a way that each respective sum of reciprocals is greater than 1. These arithmetic sequences will be disjoint.

Above each of these 4 arithmetic sequences we add a splitting node; above those we attach 8 disjoint arithmetic sequences with difference 8 which are again so long that their sum of reciprocals is each greater than 1. Etc.

Clearly any two branches of this tree will be almost disjoint sets, and along any branch the sum of reciprocals diverges.